- Dear Barukh,

The answer is yes.

Let BD, CE be the bisectors

and I the incenter of triangle ABC.

The circle (c1) (B, BD) meets BC at D'

at right of B and

the circle (c2) (C, CE) meets BC at E'

at left of C.

Since BD + CE > BI + CI > BC the point

D' is at right of E'.

When the angle B increases from 0 to 180

in order to construct ABC with BC = a and BD = t_b

the point D moves on the circle (c1) we double

the angle CBD and with line CD we find the vertex A.

The locus of A is a curve (c3) that passes through D'.

The orthogonal projection A' of A on BC moves

from D' to the left as angle B increases.

Similarly in order to construct ABC

with BC = a and CE = t_c the point E moves on

the circle (c2), the point A moves on

a curve (c4) that passes through E' and

the point A' moves from E' to the right.

Hence the curves (c3) and (c4) have only one

point of intersection (on the positive semiplane

relative to BC) and the triangle ABC

with BC = a, BD = t_b, CE = t_c is unique.

Best regards

Nikos Dergiades

>Dear Hyacinthians,

bisectors)?

>

>Is it true that there exists *at most* one triangle with

>given elements a, t_b, t_c (the last two are angle

>

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>If yes, how can we prove this uniqueness?

>

>Thanks,

>Barukh.

>

>

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- Dear Nikos,

Thank you very much for the explanations.

Barukh.

--- In Hyacinthos@y..., "Nikolaos Dergiades" <ndergiades@y...> wrote:

> Dear Barukh,

>

> The answer is yes.

>

> Let BD, CE be the bisectors

> and I the incenter of triangle ABC.

> The circle (c1) (B, BD) meets BC at D'

> at right of B and

> the circle (c2) (C, CE) meets BC at E'

> at left of C.

> Since BD + CE > BI + CI > BC the point

> D' is at right of E'.

> When the angle B increases from 0 to 180

> in order to construct ABC with BC = a and BD = t_b

> the point D moves on the circle (c1) we double

> the angle CBD and with line CD we find the vertex A.

> The locus of A is a curve (c3) that passes through D'.

> The orthogonal projection A' of A on BC moves

> from D' to the left as angle B increases.

> Similarly in order to construct ABC

> with BC = a and CE = t_c the point E moves on

> the circle (c2), the point A moves on

> a curve (c4) that passes through E' and

> the point A' moves from E' to the right.

> Hence the curves (c3) and (c4) have only one

> point of intersection (on the positive semiplane

> relative to BC) and the triangle ABC

> with BC = a, BD = t_b, CE = t_c is unique.

>

> Best regards

> Nikos Dergiades

>

> >Dear Hyacinthians,

> >

> >Is it true that there exists *at most* one triangle with

> >given elements a, t_b, t_c (the last two are angle

> bisectors)?

> >

> >If yes, how can we prove this uniqueness?

> >

> >Thanks,

> >Barukh.

> >

> >

> >------------------------ Yahoo! Groups

> Sponsor ---------------------~-->

> >Plan to Sell a Home?

> >http://us.click.yahoo.com/J2SnNA/y.lEAA/jd3IAA/6HYolB/TM

> >-----------------------------------------------------------

> ----------~->

> >

> >

> >

> >Your use of Yahoo! Groups is subject to

> http://docs.yahoo.com/info/terms/

> >